//M2//QN1//SUB//DL0//EQ

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) → N2O(g) + NO2(g); Rate = k[NO]2 (ii) H2O2(aq) + 3l–(aq) + 2H+ → 2H2O(l) + I3–; Rate = k[H2O2][I –] (iii) CH3CHO(g) → CH4(g) + CO(g);Rate = k[CH3CHO] (iv) C2H5Cl(g) → C2H4(g) + HCl(g);Rate = k[C2H5Cl]

//X

(i) 3NO(g) → N2O(g) + NO2(g)

Rate = k[NO]2

∴ Order of reaction = 2

Dimension of the rate constant:

Rate = k[NO]2

k = = = mol–1 L s–1

(ii) H2O2(aq) + 3I –(aq) + 2H+ → 2H2O(l) + I3–

Rate = k[H2O2][I –]

Order of reaction with respect to H2O2 = 1

order of reaction with respect to l–1 = 1

∴ overall order of reaction = 2

Dimension of the rate constant.

Rate = k[H2O2][I–]

k =

=

= mol–1 L s–1

(iii) CH3CHO(g) → CH4(g) + CO(g)

Rate = k[CH3CHO]

∴ Order of reaction =

Dimension of the rate constant

Rate = k[CH3CHO]

k =

=

=

(iv) C2H5Cl(g) → C2H4(g) + HCl(g)

Rate = k[C2H5Cl]

∴ Order of reaction = 1

Dimension of the rate constant.

Rate = k[C2H5Cl]

k =

= = s–1

//M3//QN2//SUB//DL0//EQ

For the reaction: 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1,

[B] = 0.2 mol L–1, Calculate the rate of reaction after [A] is reduce to 0.06 mol L–1

//X

Initial rate = k[A][B]2

= 2.0 × 10–6 mol–2 L2 s–1 × 0.1 mol L–1

× (0.2 mol L–1)2

= 8.0 × 10–9 mol L–1 s–1

Concentration of A is reduce to 0.06 mol L–1

Concentration of [A] reacted = 0.10 – 0.06

= 0.04 mol L–1

As per the reaction, 2A + B → A2B

Half of the concentration of B is reduce then A

∴ Concentration of [B] reacted = X 0.04

= 0.02 mol L–1

[B] = 0.2 – 0.02 = 0.18 mol L–1

Rate = k [A][B]2

= 2.0 × 10–6 mol–2 L2 s–1 × 0.06 mol L–1

× (0.18 mol L–1)2

= 3.88 × 10–9 mol L–1 s–1

//M2//QN3//SUB//DL0//EQ

The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10– 4

mol L–1 s–1 ?

//X

The decomposition reaction of ammonia is

2NH3(g) → N2(g) + 3H2(g),

This is zero order reaction,

Rate = =

= = k[NH3]0

Rate of appearance of N2 =

= k[NH3]0 = k

= 2.5 × 10–4 mol L–1 s–1

Rate of appearance of H2 =

= k[NH3]0 = k

∴ = 3k = 3 × 2.5 × 10– 4 mol L–1 s–1

= 7.5 × 10–4 mol L–1 s–1

//M0//QN4//SUB//DL0//EQ

The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k[CH3OCH3] The rate of reaction is followed by increase in pressure in a closed vessel. So the rate can also be expressed in terms of the partial pressure of dimethyl ether i.e. Rate = If the pressure is measured in bar and time in minutes, then what are the units of rate and rate costants ?

//X

The pressure is measured in bar and time in minutes,

So, the unit of rate = Bar min–1

Unit of K =

=

= min–1

//M0//QN5//SUB//DL0

Mention the factors that affect the rate of a chemical reaction.

//X

//M2//QN6//SUB//DL0//EQ

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i) doubled (ii) reduced to half ?

//X

Reaction → Products

Rate r1 = k[A]2

(i) When concentration of reactant is doubled;

Rate r2 = k[2A]2 = 4k[A]2 = 4r1

∴ When concentration of the reactant is doubled, the rate of reaction will become 4 times of the initial rate.

(ii) When concentration of reactant is reduced to half;

Rate r2 = = k[A]2 =

∴ when concentration of the reactant is reduced by half, the rate of reaction will be reduce to of the initial rate.

//M2//QN7//SUB//DL0//EQ

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

//X

Most of the chemical reactions are accelerated by increase in temperature.

Increasing the temperature will result in an exponential increase in the rate constant.

It has been found that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Rate = (2)

According to Swedish chemist Arrhenius, the quantitative effect of temperature on the rate constant can be express by following equation.

k = A

Where, A = Arrhenius factor or frequency factor.

R = gas constant

Ea = activation energy in terms of
joules/mole (J mol–1).

//M2//QN8//SUB//DL0//EQ

In a pseudo first order reaction in water, the following results were obtained: calculate the average rate of reaction between the time interval 30 to 60 second.

//X

Hydrolysis of ester is pseudo first order reaction.

RCOOR' + H2O RCOOH + R'OH

Ester Water Acid Alcohol

Average rate of reaction rav =

rav = –

[R]2 = concentration at 60 s = 0.17 M

[R]1 = concentration at 30 s = 0.31 M

rav =

rav = = 4.67 × 10–3 mol L–1 s–1

//M2//QN9//SUB//DL0

A reaction is first order in A and second

order in B. (i) Write the differential rate equation. [June 2025] (ii) How is the rate affected on increasing the concentration of B three times ? (iii) How is the rate affected when the concentration of both A and B are doubled ?[Topic 3.2] [June 2025] [2 Marks]

//X

//M4//QN10//SUB//DL0//EQ

In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

//X

Suppose,

Order of reaction with respect to A = x

Order of reaction with respect to B = y

Then,

Rate = k[A]x[B]y

∴ r1 = k[0.20]x[0.30]y = 5.07 ×10–5 ... ... (1)

r2= k[0.20]x[0.10]y = 5.07 ×10–5 ... ... (2)

r3 = k[0.40]x [0.05]y = 1.43 ×10– 4 ... ... (3)

Dividing (1) and (2) we get,

∴ = [3]y = 1

∴ y = 0, because [3]0 = 1 = [3]y

Dividing (3) and (2) we get,

= 2.82

∴ = 2.82

∴ [2]x = 2.82

Taking log of both the side we get,

log[2]x = log2.82

∴ x log2 = log2.82

x × 0.3010 = 0.4503

∴ x = = 1.5

∴ order of reaction with respect to A = 1.5

order of reaction with respect to B = 0

//M4//QN11//SUB//DL0//EQ

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D Determine the rate law and the rate constant for the reaction.

//X

Suppose,

Order of reaction with respect to A = x

Order of reaction with respect to B = y

∴ Rate = k[A]x[B]y

∴ r1 = k[0.1]x[0.1]y = 6.0 × 10–3 ... ... (1)

r2 = k[0.3]x[0.2]y = 7.2 × 10–2 ... ... (2)

r3 = k[0.3]x[0.4]y = 2.88 × 10–1... ... (3)

r4 = k[0.4]x[0.1]y = 2.40 × 10–2... ... (4)

Dividing (4) and (1) we get,

∴ = [4]x = [4]1

Order of reaction with respect to A = 1

Dividing (3) and (2) we get,

∴ = [2]y = 4 = [2]2

∴ y = 2

∴ order of reaction with respect to A = 1

order of reaction with respect to B = 2

overall order of reaction = 3

∴ rate equation,

Rate = k[A]1[B]2

Calculation of rate constant k

From equation (1)

r1 = k[0.1]1[0.1]2 = 6.0 × 10–3

k =

=

= 6.0 mol–2 L–2 min–1

//M3//QN12//SUB//DL0//EQ

The reaction between A and B first order with respect to A and zero order with respect to B. Fill the blanks in the following table:

//X

Reaction A → B

Rate = k[A]1[B]0

Tate constant in Experiment-I:

Rate = k[A]

∴ k =

∴ k = 2.0 × 10–1 = 0.2 min–1

[A] in Experiment-II:

Rate = k[A]

∴ [A] =

∴ [A] = 0.2 mol L–1

Initial rate in Experiment-III:

Rate = k[A]

∴ rate = 0.2 × 0.4 = 0.08 mol L–1 min–1

[A] in Experiment-IV:

Rate = k[A]

∴ [A] =

∴ [A] = 0.1 mol L–1

//M1//QN13//SUB//DL0//EQ

Calculate the half-life of a first order reaction from their rate constants given below: (1) 200 s–1 (ii) 2 min–1 (iii) 4 years–1

//X

For first order reaction =

(i) = 3.465 × 10–3 s

(ii) = 0.3465 min

(iii) = 0.17325 year

//M4//QN14//SUB//DL0//EQ

The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

//X

Half-life of 14C = 5730 years.

All radioactive decay follows first order kinetics.

k =

k = year–1

For first order reaction.

t = log

t = log

t = log1.25

t = × 0.0969

t = 1845.2 years.

//M4//QN15//SUB//DL0//EQ

The experimental data for decomposition of N2O5; [2N2O5 → 4NO2 + O2] in gas phase at 318K are given below: (i) Plot [N2O5] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log [N2O5] and t (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).

//X

(i)

(ii) Initial concentration of N2O5

= 1.63 × 10–2 M

Half of this concentration = 0.815 × 10–2 M

From graph, time corresponding to this concentration is 1400 s.

Hence, = 1400 s.

(iii) First we will find values of log [N2O5],

Time/(sec)	[N2O5] × 10+2 mol L–1	log[N2O5]
0	1.63	– 1.79
400	1.36	– 1.87
800	1.14	– 1.94
1200	0.93	– 2.03
1600	0.78	– 2.11
2000	0.64	– 2.19
2400	0.53	– 2.28
2800	0.43	– 2.37
3200	0.35	– 2.46

(iv) As plot of log [N2O5] → t is straight line. Hence, it is first order reaction.

∴ Rate = k [N2O5]

(v) From plot of log [N2O5] → t

Slope =

∴ k = – slope × 2.303

= × 2.303

= × 2.303

= 4.82 × 10–4 s–1

OR

From formula of integrated rate law for first order reaction,

k = log

∴ k = log

∴ k = log4.6571

∴ k = × 0.6681

∴ k = 4.81 × 10– 4 s–1

(vi) = =

= 0.1437 × 104 s = 1437 s

Hence, half-life from both (ii) and (vi) are nearly same.

//M3//QN16//SUB//DL0//EQ

The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value ?

//X

Rate constant k = 60 s–1

If, initial concentration = [R]0 M

Then, concentration at time t = M

For first order reaction,

t = log

= log

= log16

= × 1.2044

= 4.62 × 10-2 Sec.

//M3//QN17//SUB//DL0//EQ

During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 mg of 90Sr was absorbed in the bones of a newly born baby instead of calcium. How much of it will remain after 10 years and 60 years if it is not lost metabolically.

//X

Nuclear explosion is first order reaction.

= 28.1 years and [R]0 = 1 µg.

For first order reaction,

k = = = 2.466 × 10–2 year–1

Calculation for amount of 90Sr remain after 10 years:

t = log

∴ log =

∴ log =

∴ log = 0.1070

∴ = anti log0.1070 = 1.2794

∴ [R]t = = 0.7816 µg 90Sr is left after 10 years.

Calculation for amount of 90Sr remain after
60 years:

t = log

∴ log =

∴ log =

∴ log = 0.6425

∴ = antilog 0.6425 = 4.39

∴ [R]t = = 0.228 µg 90Sr is left after 60 years.

//M3//QN18//SUB//DL0//EQ

For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

//X

For first order reaction,

t99% = log ... ... (1)

t90% = log ... ... (2)

Taking ratio of equation (1) and (2)

=

∴ =

∴ =

∴ =

∴ = 2

∴ t99% = 2 × t90%

//M2//QN19//SUB//DL0//EQ

A first order reaction takes 40 min for 30% decomposition. Calculate the half-life period for the reaction .

//X

uppose initial concentration [R]0 = 100

∴ Concentration at time t [R]t = 70, (30% decomposition).

t = 40 min.

Comparing k = log and

k = we get,

= log

∴ = log = × 0.1549

∴ = 0.0089185

∴ = = 77.7 min

//M4//QN20//SUB//DL0//EQ

For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained. Calculate the rate constant.

//X

Decomposition reaction of azoisopropane,

(CH3)2CHN = NCH(CH3)2 → C6H14(g) + N2(g)

Pressure at time t = 0	35 mm Hg	0	0
Change in pressure	– x	+x	+x
Pressure after t = 360 s	35 – x	+x	+x

∴ Total pressure after 360 s = 35 – x + x + x = 54 mm Hg

∴ Pressure after t = 360 s; 35 – 19 mm Hg = 16 mm Hg

Rate constant at 360 s:

Initial pressure [R]0 = pi = 35 mm Hg

Pressure at time t [R]t = pt = 16 mm Hg

For first order reaction,

k = log

= log

= log2.1875

= × 0.3399

∴ k = 2.1747 × 10–3 sec–1

k at t = 720 s:

Total pressure at t = 720 s,

Total pressure at equilibrium = 35 – x + x + x

= 63 mm Hg

Initial pressure at time t = 0 s, [R]0 = pi = 35 mm Hg

Pressure at time t = 720 s, [R]t = pt = 35 – 28

= 7 mm Hg

For first order reaction,

k = log

= log

= log5

= × 0.6990

∴ k = 2.21 × 10–3 sec–1

//M4//QN21//SUB//DL0//EQ

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO2Cl2(g) → SO2(g) + Cl2(g) Calculate the rate of reaction when total pressure is 0.65 atm.

//X

Decomposition reaction of SO2Cl2,

SO2Cl2(g) → SO2(g) + Cl2(g)

Pressure at time t = 0 s pi atm 0 atm 0 atm

Pressure at time t pi – x atm x atm x atm

pi is initial pressure at time t = 0 s

∴ Total pressure at time t,

pt = pi – x + x + x = pi + x

∴ x = pt – pi

∴ p(SO2Cl2 ) = pi – x = pi – (pt – pi) = 2pi – pt atm.

For experiment - 1

At time t = 0 s, Initial pressure pi = 0.5atm

At time t = 100 s, p(SO2Cl2 ) = 2pi – pt

= 2 × 0.5 – 0.6

= 0.4 atm

For first order reaction,

∴ k = log

= log

= × 0.0969

∴ k = 2.2318 × 10–3 sec–1

For experiment-2

At time t = 0 s, Initial pressure pi = 0.5 atm

pSO2Cl2 = 2pi – pt = 2 × 0.5 – 0.65 = 0.35 atm

∴ rate = k ∙ pSO2Cl2

= 2.2318 × 0.35 × 10–3 sec–1

= 0.78113 × 10–3

= 7.8113 × 10–4 atm sec–1

//M4//QN22//SUB//DL0//EQ

The rate constant for the decomposition of N2O5 at various temperatures is given below: Draw a graph between lnk and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.

//X

Ea from the graph:

Slope =

∴ Ea = – slope × R

∴ Ea = × 8.314

∴ Ea = 94523.47 J mol-1 = 94.52347 kJ mol–1

Arrhenius constant from graph:

Intercept of graph shows value of ln A.

Intercept = 21.5

∴ ln A = 21.5

∴ 2.303 log A = 21.5

∴ log A = = 9.3356

∴ A = antilog 9.3356

∴ A = 2.1657 × 109

The rate constant at 30°C and 50°C temperature:

T = 30°C = 303K

∴ = = 3.3 × 10–3K–1

From graph, at = 3.3 × 10–3K–1, ln k = 13.8

∴ 2.303 log k = 13.8

∴ log k = = 6

∴ k = antilog 6

∴ k = 1 × 106 s–1

T = 50°C = 323K

∴ = = 3.09 × 10–3K–1

From graph, at = 3.09 × 10–3K–1, ln k = 16

∴ 2.303 log k = 16

∴ log k = = 6.9474

∴ k = antilog 6.9474

∴ k = 8.86 × 106 s-1

//M3//QN23//SUB//DL0//EQ

The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5 s–1 546K. If the energy of activation is 179.9 kJ/mol. What will be the value of pre-exponential factor.

//X

Arrhenius equation,

logk = logA –

∴ logA = logk +

Here, T = 546K, k = 2.418× 10–5 s–1

Ea = 179.9 kJ/mol = 179900 J/mol

A = ?

∴ logA = log2.418 × 10–5 +

∴ logA = log2.418 + log 10–5 + 17.2082

∴ logA = 0.3834 – 5.0 + 17.2082

∴ logA = 12.5917

∴ A = antilog12.5917 = 3.9057 × 1012 sec–1.

//M3//QN24//SUB//DL0//EQ

Consider a certain reaction A → Products with

k = 2.0 × 10–2 s–1. Calculate the concentration of A remaining after 100 sec if the initial concentration of A is 1.0 mol L–1.

//X

k = 2.0 × 10–2 sec–1

[R]0 = 1 mol L–1, t = 100 sec

[R]t = ?

k = log

∴ log =

∴ log =

∴ log = 0.8684

∴ = anti log0.8684 = 7.3863

∴ [R]t = = = 0.1354 mol L–1

//M3//QN25//SUB//DL0//EQ

Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with = 3.00 hours. What reaction of sample of sucrose remains after

8 hours?

//X

For first order reaction,

k = = = 0.231 hour–1

Fraction of sample of sucrose remains after 8 hours:

Taking initial concentration [R]0 = 1 M,

Concentration after 8 hours [R]t = ?

k = log

∴ log =

∴ log =

∴ log = 0.8024

∴ = antilog 0.8024 = 6.3445

∴ [R]t = = = 0.1576 M

∴ The fraction of sample of sucrose that remains after 8 hours is 0.1576 M.

//M3//QN26//SUB//DL0//EQ

The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s–1) . Calculate

activation energy Ea.

//X

Arrhenius equation,

k = A ... ... (I)

k = 4.5 × 1011 s–1 ... ... (II)

Taking ratio of Eq. (I) and Eq. (II) we get,

Ea = 28000K × R

Ea = 28000K × 8.314 J K–1 mol–1

Ea = 232792 J mol-1 = 232.792 kJ mol–1

//M4//QN27//SUB//DL0//EQ

The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 – 1.25 × 104 K/T Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes ?

//X

Ea is the activation energy.

According to Arrhenius equation,

k = A

log k = logA – ... ... (I)

Now, log k = 14.34 – ... ... (II)

Comparing Eq. (I) and Eq. (II) we get,

Ea = 1.25 × 104K × 2.303 × R

Ea = 1.25 × 104K × 2.303 × 8.314 J K–1 mol–1

Ea = 23.93 × 104 J mol–1 = 239.3 kJ mol–1

For first order reaction

k =

k = = 4.51 × 10–5 s–1

For given equation,

log k = 14.34 –

log 4.51 × 10-5 = 14.34 –

– 4.35 = 14.34 –

= 14.34 + 4.35 = 18.69

T = = 669K

//M3//QN28//SUB//DL0//EQ

The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104 s–1 ?

//X

A → product

Here, T1 = 10°C = 283K

k1 = 4.5 × 103 s–1

Ea = 60 kJ mol-1= 60000 J mol–1

k2 = 1.5 × 104 s–1

T2 = ?

R = 8.314 J K–1 mol–1

log

log

0.5228 =

= 0.0001668

– 0.0001668

= 0.003534 – 0.0001668 = 0.003367

T2 = = 297.00 K

T2 = 297 – 273 = 24°C

//M4//QN29//SUB//DL0//EQ

The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010 s–1. Calculate k and activation energy (Ea)at 318K.

//X

(A) Calculation of Ea:

(A) For first order reaction, k = log

At 298K temperature, k1 = log
... ... (I)

At 308K temperature, k2 = log

... ... (II)

Dividing Eq. (II) by Eq.(I) we get,

= 2.73

According to Arrhenius equation,

log

log2.73 =

Ea =

Ea = 76640 J/mol = 76.640 kJ/mol

(B) According to Arrhenius equation,

log k = logA –

log k = log4 × 1010 –

log k = 10.6021 – 12.5870 = –1.9849 = .0151

k = antilog.0151

k = 1.035 × 10–2 s–1

//M3//QN30//SUB//DL0//EQ

The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

//X

At T1 = 293K, rate constant = k1

At T2 = 313K, rate constant = k2 = 4k1

R = 8.314 J K–1 mol–1

According to Arrhenius equation,

log

log

Ea =

Ea =

Ea = 52863.33 J/mol = 52.863 kJ/mol

Class 12 Chemistry (Part 1) 013